Physics for Scientists and Engineers with Modern Physics
4th Edition
ISBN: 9780131495081
Author: Douglas C. Giancoli
Publisher: Addison-Wesley
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Chapter 37, Problem 57P
To determine
The ionization energy of doubly ionized lithium,
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=
. Using the formula for the hydrogen atom energy levels, En
constant can be written in terms of fundamental quantities,
RH
=
Me 4
8€, ²h³c
Me 4 1
860²h² n²¹
the Rydberg
and its value approaches, RH → R∞ = 10,973,731.6 m-¹ in the limit u → me.
(a) How would this constant be defined for a one-electron species containing Z protons in
its nucleus? Consider how this changes the form of the Hamiltonian and the energy
levels for that Hamiltonian.
(b) The hydrogen atom emission lines in the Balmer series (n₂ = 2) lie in the visible portion of
the electromagnetic spectrum. Would this also be true if Z> 1? Find the wavelength (in
nm) of the n = 32 emission in hydrogen and that for a one-electron species with Z = 2.
(You will be asked to report a quantity on the quiz that depends on these two values.)
Calculate the wavelength of the Mo(Z = 42)K« X-ray line given that the
ionization energy of hydrogen is 13.6 eV
[Adapted from the University of London, Royal Holloway 2002]
. (II) Estimate the wavelength for an n = 3 to n = 2
transition in iron (Z = 26).
Chapter 37 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 37.2 - Prob. 1AECh. 37.2 - Prob. 1BECh. 37.4 - Prob. 1CECh. 37.7 - Prob. 1DECh. 37.7 - Prob. 1EECh. 37.11 - Prob. 1FECh. 37 - Prob. 1QCh. 37 - Prob. 2QCh. 37 - Prob. 3QCh. 37 - Prob. 4Q
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- = Using the formula for the hydrogen atom energy levels, En constant can be written in terms of fundamental quantities, RH = Me 4 8€ ²h³c Me4 1 860²h² n²¹ the Rydberg and its value approaches, RH → R = 10,973,731.6 m¹ in the limit μ→ me. (a) How would this constant be defined for a one-electron species containing Z protons in its nucleus? Consider how this changes the form of the Hamiltonian and the energy levels for that Hamiltonian. (b) The hydrogen atom emission lines in the Balmer series (n₂ = 2) lie in the visible portion of the electromagnetic spectrum. Would this also be true if Z> 1? Find the wavelength (in nm) of the n = 32 emission in hydrogen and that for a one-electron species with Z = 2. (You will be asked to report a quantity on the quiz that depends on these two values.)arrow_forward(a) Determine the wavelength of the second Balmerline (n=4 to n=2 transition) using Fig. 27–29. Determine likewise (b) the wavelength of the second Lyman line and (c) the wavelength of the third Balmer line.arrow_forward(II) For each of the following atomic transitions, state whether the transition is allowed or forbidden, and why: (a) 4p → 3p; (b) 3p → 1s; (c) 4d → 2d; (d) 5d → 3s; (e) 4s → 2p.arrow_forward
- The Lyman series comprises a set of spectral lines. All of these lines involve a hydrogen atom whose electron undergoes a change in energy level, either beginning at the n = 1 level (in the case of an absorption line) or ending there (an emission line). The inverse wavelengths for the Lyman series in hydrogen are given by 1 - where n = 2, 3, 4, ... and the Rydberg constant R, = 1.097 x 10' m-. (Round your answers to at least one decimal place. Enter your answers in nm.) %3D (a) Compute the wavelength for the first line in this series (the line corresponding to n = 2). nm (b) Compute the wavelength for the second line in this series (the line corresponding to n = 3). nm (c) Compute the wavelength for the third line in this series (the line corresponding to n = 4). nm (d) In which part of the electromagnetic spectrum do these three lines reside? O x-ray region O ultraviolet region O infrared region O gamma ray region O visible light regionarrow_forward(a) The Lyman series in hydrogen is the transition from energy levels n = 2, 3, 4, ... to the ground state n = 1. The energy levels are given by 13.60 eV En n- (i) What is the second longest wavelength in nm of the Lyman series? (ii) What is the series limit of the Lyman series? [1 eV = 1.602 x 1019 J, h = 6.626 × 10-34 J.s, c = 3 × 10° m.s] %3D Two emission lines have wavelengts A and + A2, respectively, where AA <<2. Show that the angular separation A0 in a grating spectrometer is given aproximately by (b) A0 = V(d/m)-2 where d is the grating constant and m is the order at which the lines are observed.arrow_forward(2) Determine, the shortest and longest wavelengths of the Balmer series of hydrogen. = 1.097 x 107m-1) (RH Answerarrow_forward
- - (II) Use the result of Example 28–7 (Z = 42) to estimate the X-ray wavelength emitted when a cobalt atom (Z = 27) makes a transition from n = 2 to n = 1.arrow_forward(II) Is the use of nonrelativistic formulas justified in the Bohr atom? To check, calculate the electron's velocity, v, in terms of c, for the ground state of hydrogen, and then calculate V1 - v²/c².arrow_forwardThe wavelengths of the Lyman series for hydrogen are given by: = RH(1-1), n = 2, 3, 4, ... For the second of this series; calculate the energy (in eV). Note: 1.60 x 10^-19 J = 1.0 eV O 4.10 x 10^3 eV 2.12 x 10^3 eV 3² O 1.21 x 10^3 eV 3.30 x 10^3 eVarrow_forward
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